What is the solution for x^3y^{'''}+xy^'-y=0 ?

The solution for x^3y^{'''}+xy^'-y=0 is y=c_{1}x+c_{2}ln(x)x+c_{3}ln^2(x)x

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x^3y^{'''}+xy^'-y=0

x^{3} y^{^{'''}} + x y^{^{'}} - y = 0

y = c_{1} x + c_{2} ln (x) x + c_{3} ln^{2} (x) x

Solution steps

x^{3} y^{'''} (x) + x y^{'} (x) - y = 0

Solve linear ODE:

y = c_{1} x + c_{2} ln (x) x + c_{3} ln^{2} (x) x

y = c_{1} x + c_{2} ln (x) x + c_{3} ln^{2} (x) x

\int e^{- 1}

f (x) = \frac{2 ln ( x )}{x}

\frac{\partial}{\partial x} (2 x (4 + y)^{- 1})

\frac{d}{d x} (- 2 (e^{x} sin (x) + cos (x) e^{x}))

\int \frac{1}{5 + 2 x ^{4}} (8 x^{3}) d x

What is the solution for x^3y^{'''}+xy^'-y=0 ?
The solution for x^3y^{'''}+xy^'-y=0 is y=c_{1}x+c_{2}ln(x)x+c_{3}ln^2(x)x